FUNCTIONS OVER INTEGRAL DOMAINS OF CHARACTERISTIC ZERO
JEAN-LUC MARICHAL AND PIERRE MATHONET
Abstract. We describe the class of n-variable polynomial functions that sat-isfy Acz´el’s bisymmetry property over an arbitrary integral domain of charac-teristic zero with identity.
1. Introduction
LetR be an integral domain of characteristic zero (hence R is infinite) with iden-tity and let n⩾ 1 be an integer. In this paper we provide a complete description of all the n-variable polynomial functions overR that satisfy the (Acz´el) bisymme-try property. Recall that a function f∶ Rn → R is bisymmetric if the n2-variable mapping
(x11, . . . , x1n; . . . ; xn1, . . . , xnn) ↦ f(f(x11, . . . , x1n), . . . , f(xn1, . . . , xnn))
does not change if we replace every xij by xji.
The bisymmetry property for n-variable real functions goes back to Acz´el [1, 2]. It has been investigated since then in the theory of functional equations by several authors, especially in characterizations of mean functions and some of their extensions (see, e.g., [3, 5–7]). This property is also studied in algebra where it is called mediality. For instance, an algebra(A, f) where f is a bisymmetric binary operation is called a medial groupoid (see, e.g., [8, 9, 11]).
We now state our main result, which provides a description of the possible bisym-metric polynomial functions from Rn toR. Let Frac(R) denote the fraction field
ofR and let N be the set of nonnegative integers. For any n-tuple x = (x1, . . . , xn),
we set∣x∣ = ∑ni=1xi.
Main Theorem. A polynomial function P∶ Rn→ R is bisymmetric if and only if it is
(i) univariate, or
(ii) of degree ⩽ 1, that is, of the form P(x) = a0+ n ∑ i=1 aixi, where ai∈ R for i = 0, . . . , n, or Date: March 23, 2012.
2010 Mathematics Subject Classification. Primary 39B72; Secondary 13B25, 26B35.
Key words and phrases. Acz´el’s bisymmetry, mediality, polynomial function, integral domain. 1
(iii) of the form P(x) = a n ∏ i=1 (xi+ b)αi− b ,
where a∈ R, b ∈ Frac(R), and α ∈ Nn satisfy abk ∈ R for k = 1, . . . , ∣α∣ − 1
and ab∣α∣− b ∈ R.
The following example, borrowed from [10], gives a polynomial function of class (iii) for which b ∉ R.
Example 1. The third-degree polynomial function P∶ Z3→ Z defined on the ring
Z of integers by
P(x1, x2, x3) = 9 x1x2x3+ 3 (x1x2+ x2x3+ x3x1) + x1+ x2+ x3
is bisymmetric since it is the restriction to Z of the bisymmetric polynomial function Q∶ Q3→ Q defined on the field Q of rationals by
Q(x1, x2, x3) = 9 3 ∏ i=1 (xi+ 1 3) − 1 3.
Since polynomial functions usually constitute the most basic functions, the prob-lem of describing the class of bisymmetric polynomial functions is quite natural. On this subject it is noteworthy that a description of the class of bisymmetric lattice polynomial functions over bounded chains and more generally over distributive lat-tices has been recently obtained [4,5] (there bisymmetry is called self-commutation), where a lattice polynomial function is a function representable by combinations of variables and constants using the fundamental lattice operations∧ and ∨.
From the Main Theorem we can derive the following test to determine whether a non-univariate polynomial function P∶ Rn→ R of degree p ⩾ 2 is bisymmetric. For k∈ {p − 1, p}, let Pk be the homogenous polynomial function obtained from P by
considering the terms of degree k only. Then P is bisymmetric if and only if Ppis a
monomial and Pp(x) = P(x−b1)+b, where 1 = (1, . . . , 1) and b = Pp−1(1)/(p Pp(1)).
Note that the Main Theorem does not hold for an infinite integral domain R of characteristic r > 0. As a counterexample, the bivariate polynomial function P(x1, x2) = xr1+ xr2 is bisymmetric.
In the next section we provide the proof of the Main Theorem, assuming first thatR is a field and then an integral domain.
2. Technicalities and proof of the Main Theorem
We observe that the definition ofR enables us to identify the ring R[x1, . . . , xn]
of polynomials of n indeterminates overR with the ring of polynomial functions of n variables fromRn toR.
It is a straightforward exercise to show that the n-variable polynomial functions given in the Main Theorem are bisymmetric. We now show that no other n-variable polynomial function is bisymmetric. We first consider the special case when R is a field. We then prove the Main Theorem in the general case (i.e., whenR is an integral domain of characteristic zero with identity).
Unless stated otherwise, we henceforth assume thatR is a field of characteristic zero. Let p∈ N and let P∶ Rn → R be a polynomial function of degree p. Thus P
can be written in the form P(x) = ∑ ∣α∣⩽p cαxα, with xα= n ∏ i=1 xαi i ,
where the sum is taken over all α∈ Nn such that∣α∣ ⩽ p.
The following lemma, which makes use of formal derivatives of polynomial func-tions, will be useful as we continue.
Lemma 2. For every polynomial function B∶ Rn→ R of degree p and every x0, y0∈
Rn, we have (1) B(x0+ y0) = ∑ ∣α∣⩽p yα0 α!(∂ α xB)(x0) , where ∂xα= ∂α1 x1⋯ ∂ αn xn and α!= α1!⋯ αn!.
Proof. It is enough to prove the result for monomial functions since both sides of (1) are additive on the function B. We then observe that for a monomial function B(x) = c xβ the identity (1) reduces to the multi-binomial theorem.
As we will see, it is useful to decompose P into its homogeneous components, that is, P = ∑pk=0Pk, where
Pk(x) = ∑ ∣α∣=k
cαxα
is the unique homogeneous component of degree k of P . In this paper the homoge-neous component of degree k of a polynomial function R will often be denoted by [R]k.
Since Pp≠ 0, the polynomial function Q = P − Pp, that is
Q(x) = ∑
∣α∣<p
cαxα,
is of degree q< p and its homogeneous component [Q]q of degree q is Pq.
We now assume that P is a bisymmetric polynomial function. This means that the polynomial identity
(2) P(P(r1), . . . , P (rn)) − P(P(c1), . . . , P (cn)) = 0
holds for every n× n matrix
(3) X=⎛⎜ ⎝ x11 ⋯ x1n ⋮ ⋱ ⋮ xn1 ⋯ xnn ⎞ ⎟ ⎠∈ R n n,
where ri= (xi1, . . . , xin) and cj= (x1j, . . . , xnj) denote its ith row and jth column,
respectively. Since all the polynomial functions of degree ⩽ 1 are bisymmetric, we may (and henceforth do) assume that p⩾ 2.
From the decomposition P = Pp+ Q it follows that
P(P(r1), . . . , P (rn)) = Pp(P(r1), . . . , P (rn)) + Q(P(r1), . . . , P (rn)),
where Q(P(r1), . . . , P (rn)) is of degree p q.
To obtain necessary conditions for P to be bisymmetric, we will equate the homogeneous components of the same degree > p q of both sides of (2). By the previous observation this amounts to considering the equation
By applying (1) to the polynomial function Pp and the n-tuples x0= (Pp(r1), . . . , Pp(rn)) and y0= (Q(r1), . . . , Q(rn)), we obtain (5) Pp(P(r1), . . . , P (rn)) = ∑ ∣α∣⩽p yα0 α!∂ α xPp(x0)
and similarly for Pp(P(c1), . . . , P (cn)). We then observe that in the sum of (5) the
term corresponding to a fixed α is either zero or of degree q∣α∣ + (p − ∣α∣) p = p2− (p − q) ∣α∣
and its homogeneous component of highest degree is obtained by ignoring the com-ponents of degrees< q in Q, that is, by replacing y0 by(Pq(r1), . . . , Pq(rn)).
Using (4) with d = p2, which leads us to consider the terms in (5) for which ∣α∣ = 0, we obtain
(6) Pp(Pp(r1), . . . , Pp(rn)) − Pp(Pp(c1), . . . , Pp(cn)) = 0.
Thus, we have proved the following claim.
Claim 3. The polynomial function Pp is bisymmetric.
We now show that Pp is a monomial function.
Proposition 4. Let H∶ Rn → R be a bisymmetric polynomial function of degree p⩾ 2. If H is homogeneous, then it is a monomial function.
Proof. Consider a bisymmetric homogeneous polynomial H∶ Rn→ R of degree p ⩾ 2. There is nothing to prove if H depends on one variable only. Otherwise, assume for the sake of a contradiction that H is the sum of at least two monomials of degree p, that is,
H(x) = a xα+ b xβ+ ∑
∣γ∣=p
cγxγ,
where a b≠ 0 and ∣α∣ = ∣β∣ = p. Using the lexicographic order ≼ over Nn, we can assume that α≻ β ≻ γ. Applying the bisymmetry property of H to the n×n matrix whose(i, j)-entry is xiyj, we obtain
H(x)pH(yp) = H(y)pH(xp), where xp= (xp
1, . . . , x p
n). Regarding this equality as a polynomial identity in y and
then equating the coefficients of its monomial terms with exponent p α, we obtain
(7) H(x)p= ap−1H(xp).
SinceR is of characteristic zero, there is a nonzero monomial term with exponent (p−1) α+β in the left-hand side of (7) while there is no such term in the right-hand side since p α≻ (p − 1) α + β ≻ p β (since p ⩾ 2). Hence a contradiction.
The next claim follows immediately from Proposition 4. Claim 5. Pp is a monomial function.
By Claim 5 we can (and henceforth do) assume that there exist c∈ R ∖ {0} and γ∈ Nn, with∣γ∣ = p, such that
A polynomial function F∶ Rn→ R is said to depend on its ith variable xi (or xi
is essential in F ) if ∂xiF ≠ 0. The following claim shows that Pp determines the
essential variables of P .
Claim 6. If Pp does not depend on the variable xj, then P does not depend on xj.
Proof. Suppose that ∂xjPp= 0 and fix i ∈ {1, . . . , n}, i ≠ j, such that ∂xiPp/= 0. By
taking the derivative of both sides of (2) with respect to xij, the(i, j)-entry of the
matrix X in (3), we obtain
(9) (∂xiP)(P(r1), . . . , P(rn))(∂xjP)(ri) = (∂xjP)(P(c1), . . . , P (cn))(∂xiP)(cj).
Suppose for the sake of a contradiction that ∂xjP ≠ 0. Thus, neither side of (9)
is the zero polynomial. Let Rj be the homogeneous component of ∂xjP of highest
degree and denote its degree by r. Since Pp does not depend on xj, we must have
r< p − 1. Then the homogeneous component of highest degree of the left-hand side in (9) is given by
(∂xiPp)(Pp(r1), . . . , Pp(rn)) Rj(ri)
and is of degree p(p − 1) + r. But the right-hand side in (9) is of degree at most rp+ p − 1 = (r + 1)(p − 1) + r < p(p − 1) + r, since r < p − 1 and p ⩾ 2. Hence a
contradiction. Therefore ∂xjP= 0.
We now give an explicit expression for Pq = [P − Pp]q in terms of Pp. We first
present an equation that is satisfied by Pq.
Claim 7. Pq satisfies the equation
(10) n ∑ i=1 Pq(ri)(∂xiPp)(Pp(r1), . . . , Pp(rn)) = n ∑ i=1 Pq(ci)(∂xiPp)(Pp(c1), . . . , Pp(cn))
for every matrix X as defined in (3).
Proof. By (6) and (8) we see that the left-hand side of (4) for d = p2 is zero.
Therefore, the highest degree terms in the sum of (5) are of degree p2− (p − q) > p q
(because (p − 1)(p − q) > 0) and correspond to those α ∈ Nn for which ∣α∣ = 1.
Collecting these terms and then considering only the homogeneous component of highest degree (that is, replacing Q by Pq), we see that the identity (4) for d =
p2− (p − q) is precisely (10). Claim 8. We have (11) Pq(x) = Pq(1) c p Pp(x) n ∑ j=1 γj xp−qj .
Moreover, Pq= 0 if there exists j ∈ {1, . . . , n} such that 0 < γj< p − q.
Proof. Considering Eq. (10) for a matrix X such that ri = x for i = 1, . . . , n, we
obtain c p Pq(x) Pp(x)p−1= Pq(1) n ∑ i=1 xqi(∂xiPp)(c x p 1, . . . , c x p n).
Since ∂xiPp(x) = γiPp(x)/xi, the previous equation becomes
(12) c p Pq(x) Pp(x)p−1= Pq(1) Pp(x)p n ∑ i=1 γi xp−qi
from which Eq. (11) follows. Now suppose that Pq ≠ 0 and let j ∈ {1, . . . , n}.
Comparing the lowest degrees in xj of both sides of (12), we obtain
(p − 1) γj⩽⎧⎪⎪⎨⎪⎪
⎩
p γj− p + q , if γj≠ 0,
p γj, if γj= 0.
Therefore, we must have γj= 0 or γj⩾ p−q, which ensures that the right-hand side
of (11) is a polynomial.
If ϕ∶ R → R is a bijection, we can associate with every function f∶ Rn → R its conjugate ϕ(f)∶ Rn→ R defined by
ϕ(f)(x1, . . . , xn) = ϕ−1(f(ϕ(x1), . . . , ϕ(xn))).
It is clear that f is bisymmetric if and only if so is ϕ(f). We then have the following fact.
Fact 9. The class of n-variable bisymmetric functions is stable under the action of conjugation.
Since the Main Theorem involves polynomial functions over a ring, we will only consider conjugations given by translations ϕb(x) = x + b.
We now show that it is always possible to conjugate P with an appropriate translation ϕb to eliminate the terms of degree p− 1 of the resulting polynomial
function ϕb(P).
Claim 10. There exists b∈ R such that ϕb(P) has no term of degree p − 1.
Proof. If q< p − 1, we take b = 0. If q = p − 1, then using (1) with y0= b1, we get
[ϕb(P)]p−1= Pp−1+ b n
∑
i=1
∂xiPp.
On the other hand, by (11) we have Pp−1= Pp−1(1) c p n ∑ i=1 ∂xiPp.
It is then enough to choose b= −Pp−1(1)/(c p) and the result follows.
We can now prove the Main Theorem for polynomial functions of degree⩽ 2. Proposition 11. The Main Theorem is true when R is a field of characteristic zero and P is a polynomial function of degree ⩽ 2.
Proof. Let P be a bisymmetric polynomial function of degree p⩽ 2. If p ⩽ 1, then P is in class(ii) of the Main Theorem. If p = 2, then by Claim 10 we see that P is (up to conjugation) of the form P(x) = c2xixj+ c0. If i = j, then by Claim 6
we see that P is a univariate polynomial function, which corresponds to the class (i). If i /= j, then by Claim 8 we have c0= 0 and hence P is a monomial (up to
conjugation).
By Proposition 11 we can henceforth assume that p⩾ 3. We also assume that P is a bivariate polynomial function. The general case will be proved by induction on the number of essential variables of P .
Proposition 12. The Main theorem is true whenR is a field of characteristic zero and P is a bivariate polynomial function.
Proof. Let P be a bisymmetric bivariate polynomial function of degree p⩾ 3. We know that Pp is of the form Pp(x, y) = c xγ1yγ2. If γ1γ2= 0, then by Claim 6 we
see that P is a univariate polynomial function, which corresponds to the class(i). Conjugating P , if necessary, we may assume that Pp−1 = 0 (by Claim 10) and
it is then enough to prove that P = Pp (i.e., Pq = 0). If γ1= 1 or γ2= 1, then the
result follows immediately from Claim 8 since p− q ⩾ 2. We may therefore assume that γ1⩾ 2 and γ2⩾ 2. We now prove that P = Pp in three steps.
Step 1. P(x, y) is of degree ⩽ γ1in x and of degree⩽ γ2 in y.
Proof. We prove by induction on r∈ {0, . . . , p − 1} that Pp−r(x, y) is of degree ⩽ γ1
in x and of degree ⩽ γ2 in y. The result is true by our assumptions for r = 0
and r= 1 and is obvious for r = p. Considering Eq. (4) for d = p2− r > p q, with r1= r2= (x, y), we obtain
(13) [P(x, y)p]p2−r= [P(x, x)
γ1P(y, y)γ2]
p2−r.
Clearly, the right-hand side of (13) is a polynomial function of degree⩽ p γ1 in x
and⩽ p γ2in y. Using the multinomial theorem, the left-hand side of (13) becomes
[P(x, y)p] p2−r= [( p ∑ k=0 Pp−k(x, y)) p ] p2−r = ∑ α∈Ap,r (αp)∏p k=0 Pp−k(x, y)αk, where Ap,r= {α = (α0, . . . , αp) ∈ Np+1∶ p ∑ k=0 k αk = r, ∣α∣ = p}.
Observing that for every α∈ Ap,r we have αk= 0 for k > r and αr/= 0 only if αr= 1
and α0= p − 1, we can rewrite (13) as
p Pp(x, y)p−1Pp−r(x, y) = [P(x, x)γ1P(y, y)γ2]p2−r− ∑ α∈Ap,r αr=⋯=αp=0 (αp)r−1∏ k=0 Pp−k(x, y)αk.
By induction hypothesis, the right-hand side is of degree⩽ p γ1 in x and of degree
⩽ p γ2 in y. The result then follows by analyzing the highest degree terms in x and
y of the left-hand side.
Step 2. P(x, y) factorizes into a product P(x, y) = U(x) V (y). Proof. By Step 1, we can write
P(x, y) =
γ1
∑
k=0
xkVk(y) ,
where Vk is of degree ⩽ γ2 and Vγ1(y) = ∑
γ2
j=0cγ2−jy
j, with c
0 = c /= 0 and c1 = 0
(since Pp−1= 0). Equating the terms of degree γ21 in z in the identity
P(P(z, t), P(x, y)) = P(P(z, x), P(t, y)) , we obtain Vγ1(t) γ1V γ1(P(x, y)) = Vγ1(x) γ1V γ1(P(t, y)).
Equating now the terms of degree γ1γ2in t in the latter identity, we obtain
(14) cγ1V
γ1(P(x, y)) = c Vγ1(x)
γ1V
γ1(y)
γ2.
We now show by induction on r∈ {0, . . . , γ1} that every polynomial function Vγ1−r
To do so, we equate the terms of degree γ1γ2− r in x in (14) (by using the explicit
form of Vγ1 in the left-hand side). Note that terms with such a degree in x can
appear in the expansion of Vγ1(P(x, y)) only when P(x, y) is raised to the highest
power γ2 (taking into account the condition c1= 0 when r = γ1). Thus, we obtain
cγ1+1 [( γ1 ∑ k=0 xγ1−kV γ1−k(y)) γ2 ] γ1γ2−r = c [Vγ1(x) γ1] γ1γ2−rVγ1(y) γ2,
(here the notation[⋅]γ1γ2−rconcerns only the degree in x). By computing the
left-hand side (using the multinomial theorem as in the proof of Step 1) and using the induction on r, we finally obtain an identity of the form
a Vγ1(y) γ2−1V γ1−r(y) = a ′ Vγ1(y) γ2, a, a′∈ R, a ≠ 0,
from which the result immediately follows.
Step 3. U and V are monomial functions.
Proof. Using (14) with P(x, y) = U(x) V (y) and Vγ1= V , we obtain
(15) cγ1 γ2 ∑ j=0 cγ2−j(U(x) V (y)) j= c V (x)γ1V(y)γ2.
Equating the terms of degree γ2
2 in y in (15), we obtain (16) cγ1+γ2+1U(x)γ2= cγ2+1V(x)γ1. Therefore, (15) becomes γ2−1 ∑ j=0 cγ2−j(U(x) V (y)) j= 0,
which obviously implies ck= 0 for k = 1, . . . , γ2, which in turn implies V(x) = c xγ2.
Finally, from (16) we obtain U(x) = xγ1.
Steps 2 and 3 together show that P = Pp, which establishes the proposition.
Recall that the action of the symmetric group Sn on functions fromRn toR is
defined by
σ(f)(x1, . . . , xn) = f(xσ(1), . . . , xσ(n)), σ∈ Sn.
It is clear that f is bisymmetric if and only if so is σ(f). We then have the following fact.
Fact 13. The class of n-variable bisymmetric functions is stable under the action of the symmetric group Sn.
Consider also the following action of identification of variables. For f∶ Rn→ R
and i< j ∈ [n] we define the function Ii,jf∶ Rn−1→ R as
(Ii,jf)(x1, . . . , xn−1) = f(x1, . . . , xj−1, xi, xj, . . . , xn−1).
This action amounts to considering the restriction of f to the “subspace of equation xi= xj” and then relabeling the variables. By Fact 13 it is enough to consider the
identification of the first and second variables, that is,
(I1,2f)(x1, . . . , xn−1) = f(x1, x1, x2. . . , xn−1).
Proposition 14. The class of n-variable bisymmetric functions is stable under identification of variables.
Proof. To see that I1,2f is bisymmetric, it is enough to apply the bisymmetry of f to the n× n matrix ⎛ ⎜⎜ ⎜ ⎝ x1,1 x1,1 ⋯ x1,n−1 x1,1 x1,1 ⋯ x1,n−1 ⋮ ⋮ ⋱ ⋮ xn−1,1 xn−1,1 ⋯ xn−1,n−1 ⎞ ⎟⎟ ⎟ ⎠ .
To see that Ii,jf is bisymmetric, we can similarly consider the matrix whose rows
i and j are identical and the same for the columns (or use Fact 13). We now prove the Main Theorem by using both a simple induction on the number of essential variables of P and the action of identification of variables.
Proof of the Main Theorem whenR is a field. We proceed by induction on the num-ber of essential variables of P . By Proposition 12 the result holds when P depends on 1 or 2 variables only. Let us assume that the result also holds when P depends on n− 1 variables (n − 1 ⩾ 2) and let us prove that it still holds when P depends on n variables. By Proposition 11 we may assume that P is of degree p⩾ 3. We know that Pp(x) = c xγ, where c≠ 0 and γi > 0 for i = 1, . . . , n (cf. Claim 6). Up to a
conjugation we may assume that Pp−1= 0 (cf. Claim 10). Therefore, we only need
to prove that P = Pp. Suppose on the contrary that P− Pp has a polynomial
func-tion Pq≠ 0 as the homogeneous component of highest degree. Then the polynomial
function I1,2P has n−1 essential variables, is bisymmetric (by Proposition 14), has
I1,2Ppas the homogeneous component of highest degree (of degree p⩾ 3), and has
no component of degree p−1. By induction hypothesis, I1,2P is in class(iii) of the
Main Theorem with b= 0 (since it has no term of degree p − 1) and hence it should be a monomial function. However, the polynomial function[I1,2P]q= I1,2Pq is not
zero by (11), hence a contradiction.
Proof of the Main Theorem whenR is an integral domain. Using the identification of polynomials and polynomial functions, we can extend every bisymmetric poly-nomial function over an integral domainR with identity to a polynomial function on Frac(R). The latter function is still bisymmetric since the bisymmetry prop-erty for polynomial functions is defined by a set of polynomial equations on the coefficients of the polynomial functions. Therefore, every bisymmetric polynomial function overR is the restriction to R of a bisymmetric polynomial function over Frac(R). We then conclude by using the Main Theorem for such functions.
Acknowledgments
The authors wish to thank J. Dasc˘al and E. Lehtonen for fruitful discussions. This research is supported by the internal research project F1R-MTH-PUL-12RDO2 of the University of Luxembourg.
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Mathematics Research Unit, FSTC, University of Luxembourg, 6, rue Coudenhove-Kalergi, L-1359 Luxembourg, Luxembourg
E-mail address: jean-luc.marichal[at]uni.lu
University of Li`ege, Department of Mathematics, Grande Traverse, 12 - B37, B-4000 Li`ege, Belgium
